Freaky Fraction Cancelling

As a follow up to my last post, take the example of ‘cancelling,’ a common technique in algebra. A student may be asked to simplify the fraction:

\frac{13 \times 20}{20}

Instead of actually calculating 13 times 20, and then dividing by 20, a clever student may instead ‘cancel’ the 20 on the top and the bottom giving a final answer of 13. This can also be done with more complicated fractions. For example:

\frac{13 \times 50}{25}

Since 50 = 2 \times 25, we can cancel the 25 on the top and bottom to leave us with the much simpler expression: 13 \times 2

However, my students often do not understand the limits of this cancelling technique. For example, they may think that in the fraction


the 3’s can be cancelled and reduce the fraction to


This leads to a whole mess of confusions.

Instead, we will start with the statement ‘you can always cancel any number on the top and bottom of a fraction.’

Next, a few choice examples make this rule seem plausible:



We can check that each example is valid by comparing the decimal representations on a calculator. Sure enough, the initial and final fractions are the same. The students now think that the rule is true. While they have not proved it, emotionally, they are sold on the technique. They believe it.

It is at this moment we create a shocking counterexample:


Everyone knows that 2 is not equal to 4. But our method has produced just that. Thus, our method is incorrect. The statement ‘you can always cancel any number on the top and bottom of a fraction’ must be false.

Of course you can sometimes cancel numbers and get the correct answer. But math is not about sometimes. If we state a general rule, it should hold all the time. By exhibiting a counterexample where the rule fails, and doing so with an emotional shock, hopefully students will not attempt to follow the misguided rule.

Leave a comment

Filed under Uncategorized

All Sheep are White

Counterexamples. They are a cornerstone of mathematical discourse. Normally, a mathematician attempts to produce a counterexample when he or she is uncertain about the truth of a statement.

For example, the statement ‘all sheep are white’ might seem on the surface to be true. As one looks online for pictures of sheep, the results all return white sheep:pexels-photo-288621.jpeg

This gives us some initial confidence that the statement may be true. We are beginning to believe in the statement ‘all sheep are white.’ However, and this is crucial, we have not proven it. Emotionally, we may believe it, but logically, we are not certain.

It is at this point that we must switch gears in our mathematical thinking. Instead of searching for more pictures of white sheep, we now try to disprove the statement. This is done by producing a counterexample; a sheep that is not white.

Maybe we try a search for ‘black sheep.’ Sure enough, after a bit of poking around on the internet, we find a sheep that is not white!pexels-photo-372672.jpeg

We have disproven the statement that ‘all sheep are white.’ This type of back and forth reasoning is essential when thinking mathematically. The black sheep we found served as the counterexample to our statement and settled the issue for us.

In my teaching experience, I have often found that my students do not appreciate the value of counterexamples. I will often show them a counterexample to a statement and most of them simply shrug.

My theory is that my students have not been convinced to believe in the initial statement. In our sheep example, after seeing endless pictures of white sheep, the sudden appearance of a black sheep creates an emotional response. We were convinced all sheep were white; we believed the initial statement.

As teachers, we need to ensure that the initial statement has been shown to be plausible before we demolish it with a counterexample. Consider the follow:

How many regions are inside this circle?


Let’s add another point. How about now?


And now?


You probably don’t even need to count on the next one:


I bet you noticed this pattern:

2, 4, 8, 16, …

What do you think is next?

Go ahead, count it up:







If you are having trouble, here are some labels:


The pattern is 2, 4, 8, 16, 31, 57, 99. Surprising hey? If you are like me, after the 2, 4, 8, you probably suspected the next count would be 16. When that was confirmed you believed that the next number would be 32. You had no proof, but emotionally it just seemed right.

This is why the 31 had an emotional impact. The ‘black sheep’ was a suprise, a shock. It wasn’t just a random counterexample.

The take away for education is that we, as teachers, need to ‘sell’ our students on the initial statement before we show the counterexample. We need to convince them that is all sheep might actually be white, before we show them a black one.


1 Comment

Filed under Uncategorized

Word of the Day – Isomorphic

Let’s play a game. First, you need an opponent. Next, you each take turns selecting a number from the spinner below. The first person to collect three numbers that add up to 12 wins.

Here is an example game:

Person A chooses 4. Person B chooses 7. Person A chooses 6. Person B chooses 3. Person A chooses 2 and wins, since 4 + 6 + 2 = 12

Give the game a try yourself to get the feel of it. You should realize that it is a fairly challenging game. Often, you get a couple number at the start, but are then forced to ‘block’ your opponent. Personally, a large number of my games ended in a draw because no one was able to get the right numbers to sum to 12.

As in any game, we want to develop a winning strategy. To start, we could write down all the sets of three numbers that add up to 12:

  • 0 + 4 + 8
  • 0 + 5 + 7
  • 1 + 3 + 8
  • 1 + 4 + 7
  • 1 + 5 + 6
  • 2 + 3 + 7
  • 2 + 4 + 6
  • 3 + 4 + 5

Next, we notice that the number 4 shows up in four of the sums, while the other numbers only show up two or three times. Hence, choosing 4 first would be a reasonable move. From there, we could try to keep our list of sums handy and use it to our wit our opponent. However, this strategy is cumbersome at best, and does not seem very intuitive. Perhaps there is a better way to understand the game…

The study of mathematics involves discovering structure and pattern. Frequently, a mathematician may find two objects that appear to have a similar structure or form. In our case, we wish to find a second object that has a similar structure to our spinner game. Since we need to add up three of the nine total numbers, we could consider a 3 X 3 grid:

Now we need to place our nine numbers somewhere in the grid. Since we want winning combinations, we should arrange all nine numbers in the grid so that every row, column, and diagonal adds up to 12 (our 8 different sums we listed above). Since the middle square is involved in four different sums (down, across, left diagonal, right diagonal) we place our most common number, the number 4, in the middle:

After some guessing and checking, we find the following placements of numbers appear to work for the diagonal sums:

We fill in the remaining spaces with our remaining numbers, ensuring the sum of each column and row is always 12. Voila:

For those of you who are interested, this type of construction has a special name, a magic square.

Back to our game. Now that we have our spinner in a different form, we can try using the grid to help us determine some strategy. Let’s represent Player A’s choice with an X and Player B’s choice with an O. Here is the first game from the beginning of the post:

Hmm, I think I have seen this game before… When a mathematician finds two things that have the same structure or form, they use a special word. That word is Isomorphic, which originates from the Greek iso, meaning “equal,” and morphic, meaning “shape” or “form.”

Due to our investigation above, we can confidently state that the spinner game and Tic-tac-toe are Isomorphic. This revelation was surprising to me. The spinner game seemed complicated and required a lot of thinking. Tic-tac-toe, on the other hand, is a child’s game that I mastered a long time ago. The fact that these two games are isomorphic shows just how useful looking at a problem from a different perspective can be.

1 Comment

Filed under Uncategorized

The Monty Hall Problem

Step right up, step right up! Try your luck at our fabulous contest. Guess right and you win a new car! Guess wrong, and you have to take home a goat!

You’re given the choice of three doors: Behind one door is a car. Behind the other two, smelly goats! You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3. He shows you that door No. 3 has a goat. He then says to you, “Do you want to switch to door No. 2?” Is it to your advantage to switch your choice?

Many people want to stay with their choice. This is because their gut reaction told them to pick door No. 1 and there is no reason to switch now. In fact, some contestants may even be suspicious of the game show hosts intentions. How do they know he won’t trick them and have them switch to a door with a goat!? A mathematical analysis is needed if clear thinking is to prevail.

As in similar situations, we need to break down this problem in to cases. To simplify the analysis, we will assume the car is always behind Door No. 1. This is because picking any other door and then doing a case by case analysis would be symmetric no matter what the numberings.

Case 1 (Pick Door No. 1):

\begin{array}{|c|c|c|}\hline  \textbf{Door No. 1} & \textrm{Door No. 2} & \textrm{Door No. 3} \\ \hline  \textrm{Car} & \textrm{Goat} & \textrm{ Goat}  \\ \hline \end{array}

Suppose you picked the door with the car, Door No. 1. But you don’t know it yet. And worse, the game show host is now opens door No. 2 and shows you a goat. He suggests you switch to door No. 3. If you switch, you will get a goat and lose. If you stay, you win a new car! Hmm, sticking with your original choice seems ideal in this situation.

Case 2 (Pick Door No. 2):

\begin{array}{|c|c|c|}\hline  \textrm{Door No. 1} & \textbf{Door No. 2} & \textrm{Door No. 3} \\ \hline  \textrm{Car} & \textrm{Goat} & \textrm{ Goat}  \\ \hline \end{array}

Now you picked a door with a goat, yuk! The host opens door No. 3 and shows you a goat. If you switch to door No. 1, you win the car. If you stay, you “win” the goat, not a good plan.

Case 3 (Pick Door No. 3):

\begin{array}{|c|c|c|}\hline  \textrm{Door No. 1} & \textrm{Door No. 2} & \textbf{Door No. 3} \\ \hline  \textrm{Car} & \textrm{Goat} & \textrm{ Goat}  \\ \hline \end{array}

Last case, you again have a door with a goat. The host shows you door No. 2 with a goat. If you switch to door No. 3 you win the car. If you stay, sadly, you will go home with the goat. Now let’s evaluate the strategies.

Say you are convinced that your gut is always right, and you never change your door. In case 1 you won. But your paranoia caused you to lose out big time in case 2 and 3. Hence, your chance of winning with the “stay” strategy is 1/3.

What if you are an indecisive person and you always change your mind (and your door) whenever possible. In case 1, you would change away from the car and get a goat. But in cases 2 and 3 your indecisiveness gave you the victory. Hence, your chance of winning with the “change” strategy is 2/3.

We can see that, overall, the better strategy is to change to the other door. This should be clear by the analysis above, and if you don’t believe me, try it for yourself next time you are on a game show.

However, if you would indulge me for just a moment, I want to muse on an interpretation of these results. If you pick any door at random, your chance of getting the car is 1/3. Next, the host opens a door to show you a goat. The math above states that, just by opening that door, the game has radically changed. If you stay with your door, you have a 1/3 chance of winning, just like before. But if you switch doors, suddenly your chance of winning jumps to 2/3!

But think about it logically. If there are only 2 doors left, and one has a goat and one has a car, shouldn’t your probability be 50-50, 1/2 chance either if you stay with your door or if you switch?

This is where the mystery of the problem comes in. Most people (wrongly) assume that since there are 2 doors left and 1 car they have a 50% chance of winning. So they see no reason to switch doors. In reality however, switching increases their chances of winning from 33% to 67%. But you protest, originally all 3 doors just had a 33% chance of winning, what in the world is going on?

The best I can explain it is this. When the game show host opened a door, he removed that door from the game. The 33% probability that that door was a winner had to go somewhere. And it did. This probability “collapsed” into the probability of the other door, giving the “switch” door a probability of 67% compared with your current “stay” door probability of 33%. Mind blowing. If you want the car, switch your door. If you want a goat, just say nope 🙂

Leave a comment

Filed under Uncategorized

When it Doesn’t Distribute

One of the things I, and probably all math teachers, struggle with is teaching students when they can and cannot distribute. For example, students learn very early on in their mathematical education that:

3(4+8) = 3\times 4 + 3 \times 8 = 12 + 24 = 36

One reason we know this is valid is because we can calculate the answer using the rules of BEDMAS and obtain the same final answer:

3(4+8) = 3(12) = 36

This pattern extends to variables as well. Often it takes students a few tries, but eventually they become proficient in questions like:

5(x+2) = 5x + 10

However, the problem begins in grade 10. Students begin to encounter expressions like (x+3)^2. They believe that the same pattern should hold. And so, they dutifully write:

(x+3)^2 = x^2 + 3^2 = x^2+9

Next, they apply this pattern of distributing incorrectly to square roots:

\sqrt{9+16} = \sqrt{9} + \sqrt{16} = 3 + 4 = 7

Where clearly the answer should be:

\sqrt{9+16} = \sqrt{25} = 5

And once grade 11 and 12 hit, the number of bad distributing examples proliferates!

(x+2)^3 = x^3 + 2^3 = x^2+8

\sin (x+y) = \sin x + \sin y

\log{(x+10)} = \log x + \log 10

5(xy) = 5x5y

Ultimately, I think we as math teachers are to blame for the problem. Since distributing is so natural for us, we gloss over how unique and remarkable it actually is. As such, we give students a cavalier attitude towards distributing and they happily apply it in every context they encounter. As the above list suggests, distributing is very rare and only applies in a vary narrow set of situations. When teaching the distributive property, we should do a better job demonstrating just how unusual this property is.



1 Comment

Filed under Uncategorized

Complex Connections

The other day I was investigating a problem involving complex numbers when I was hit by a flash of insight. Here is the problem as it was originally stated:

If z is a complex number such that |z| = 1, compute:

|1-z|^2 + |1+z|^2

Initially, the problem seemed very dry. My first solution involved multiplying out a bunch of terms, combining like terms, and arriving at an answer of 4. However, as I was drawing a diagram of the problem, something amazing happened.

I realized that the expression |1-z| could be interpreted as the distance from the point (1,0) to the number . But the constraint that |z| = 1,  meant z had to lie on a circle of radius 1 centered at the origin.

Further, the expression |1+z| = |1-(-z)|  could be interpreted as the distance from the point (1,0) to the number -z, which would be directly opposite the original z .

From here, all I had to do was connect the dots and observe the solution. I had formed a right-triangle! If you want to play around with the diagram, feel free to click on the interactive link:

Hence, the expression: |1-z|^2 + |1+z|^2 , could be rewritten as:

(\textrm{side}_1)^2 + (\textrm{side}_2)^2

And using Pythagoras’s theorem, we get:

a^2 + b^2 = c^2

Now the hypotenuse is the diameter of the circle. Since the circle has radius 1, the diameter is 2, and:

|1-z|^2 + |1+z|^2 = 2^2 = 4

The question ultimately boiled down to using Pythagoras’s theorem. I continue to be amazed at how many times Pythagoras’ theorem is used. A simple relationship, taught to students in grade 7 or 8, continues to be relevant when evaluating complex valued expressions. Remarkable.

Leave a comment

Filed under Uncategorized

Multiple Choice vs Matching Part 2

In the last post, we discovered that random guessing results in the same average score on both a matching question and a set of multiple choice questions. However, I still feel as though there is a difference between the two types of questions. Personally, the fact that it is impossible to score a 3 on the matching question piques my interest. To investigate the problem, we can use some of the results from the previous post.

Recall, we had the following table for the matching problem:

\begin{array}{|c|c|c|c|}\hline 0 \textrm{ Points} & 1 \textrm{ Point} & 2 \textrm{ Points}& 4 \textrm{ Points}\\ \hline 9 \textrm{ Ways} & 8 \textrm{ Ways} & 6 \textrm{ Ways} & 1 \textrm{ Way} \\ \hline  \end{array}

We can convert the ‘ways’ into probabilities by dividing by the total number of ways, 24. This gives:

\begin{array}{|c|c|c|c|c|}\hline 0 \textrm{ Points} & 1 \textrm{ Point} & 2 \textrm{ Points} & 3 \textrm{ Points}& 4 \textrm{ Points}\\ \hline 37.5\% & 33.3\%  & 25 \% & 0 \% & 4.2\% \\ \hline  \end{array}

Next, we need to compute the same table for the multiple choice situation. The math is simple, if a bit long:

\begin{array}{|c|c|c|c|c|}\hline 0 \textrm{ Points} & 1 \textrm{ Point} & 2 \textrm{ Points} & 3 \textrm{ Points}& 4 \textrm{ Points}\\ \hline 31.6\% & 42.2\%  & 21.1 \% & 4.7 \% & 0.4\% \\ \hline  \end{array}

That table is strange indeed. Comparing the two types of questions, nothing looks the same between them! Starting on the left, we see that with the matching question, you are around 6% more likely to score no points when compared to the multiple choice questions. That’s 6 more students out of 100 who would score no marks! Then, on the far right, you are 10 times more likely to score 4 points (full marks) on the matching than on the multiple choice! If you had a class of 100, 4 would score full marks on the matching while none would score full marks on the multiple choice. That hardly seems fair.

In fact, it feels like the matching question is ‘all or nothing.’ While you might score 1 point on average (see the previous post), you are more likely to score no points or full points than if you had been guessing on a set of multiple choice questions.

In statistics, we quantify this variability in the outcome using the standard deviation. Using this language, we would say our matching question has a higher standard deviation than our multiple choice question, even though they have the same average.

As an analogy, consider a room of big NFL players and skinny mathematicians. The average weight of a human in the room is probably around 200 pounds. However, there will be a huge variability in that weight; with some people weighing in around 275 and others around 150. This means our room would have a high standard deviation.

If instead, you had a room consisting of only 5 foot 10-inch-tall males with an average build, the average weight might still be around 200 pounds. But most of these people would be close to the average weight. This means our second room would have a low standard deviation.

So which question type (or room) is fairer? It really comes down to preference. While having question types with low variability might seem ideal, this is not set in stone. We appreciate the variability in the human population. If everyone was basically the same size, football wouldn’t be nearly as interesting. In fact, taking the low variability idea to the extreme, we could create a test that consisted entirely of true-false questions. While this test might seem fairer, I doubt many students would enjoy it. I for one, appreciate variability in my test questions.

Leave a comment

Filed under Uncategorized